Answer:the mass of water vapor is=d. 2.04 kg
Step-by-step explanation:
The humidity ratio by mass is given as
x = mw / ma (1)
where
mw = mass of water vapor
ma = mass of dry air
And also humidity ratio by pressure by Ideal Gas Law is given as
x = 0.62198 pw / (pa - pw) (2)
where
pw = partial pressure of water vapor in moist air
pa = atmospheric pressure of moist air
Equating both equation we have that
Humidity ratio = 0.62198 pw / (pa - pw)= mw / ma (3)
From the from the water table at 25°C,
water vapor partial pressure = 23.8 torr
1 atm =760 torr.
1 atm = 101.325 kPa,
So, 760 torr =101.325 kPa,
Therefore 23.8 torr =3.173kPa
Humidity ratio = 0.62198 pw / (pa - pw)= mw / ma
0.62198 x 3.173 / (100-3.173)= mw / 100
1.9736 /96.827=mw/100
0.0204=mw/100
mw=2.04kg